Question-49183

Question Number 49183 by ajfour last updated on 04/Dec/18

Commented by ajfour last updated on 04/Dec/18

F i n d \frac{A r e a △ P Q R}{A r e a △ A B C}, i f c e n t r o i d G

o f △ A B C l i e s o n z a x i s a n d e a c h

v e r t e x o n s u r f a c e o f p a r a b o l o i d .

Commented by mr W last updated on 04/Dec/18

d o e s Δ A B C h a v e g i v e n s i d e l e n g t h e s

a, b, c ?

Commented by ajfour last updated on 04/Dec/18

y e s s i r!

Answered by mr W last updated on 05/Dec/18

z = {k r}^{2} w i t h k = 1

l e n g t h e s o f m e d i a n s o f Δ A B C :

A G = \frac{2}{3} m_{a} = g_{a} = \frac{\sqrt{2 (b^{2} + c^{2}) - a^{2}}}{3}

B G = \frac{2}{3} m_{b} = g_{b} = \frac{\sqrt{2 (c^{2} + a^{2}) - b^{2}}}{3}

C G = \frac{2}{3} m_{c} = g_{c} = \frac{\sqrt{2 (a^{2} + b^{2}) - c^{2}}}{3}

l e t z_{G} = h

l e t z_{A} = z_{G} + A G \sin α = h + g_{a} \sin α

l e t z_{B} = z_{G} + B G \sin β = h + g_{b} \sin β

w e h a v e t h r e e u n k n o w n s : h, α, β .

r_{A} = A G \cos α = g_{a} \cos α

r_{B} = B G \cos β = g_{b} \cos β

z_{A} = {k r}_{A}^{2}

\Rightarrow h + g_{a} \sin α = {k g}_{a}^{2} \cos^{2} α

\Rightarrow \sin^{2} α + \frac{1}{{k g}_{a}} \sin α + \frac{h}{{k g}_{a}^{2}} - 1 = 0 \dots (i)

s i m i l a r l y

\Rightarrow \sin^{2} β + \frac{1}{{k g}_{b}} \sin β + \frac{h}{{k g}_{b}^{2}} - 1 = 0 \dots (i i)

\Rightarrow \sin^{2} γ + \frac{1}{{k g}_{c}} \sin γ + \frac{h}{{k g}_{c}^{2}} - 1 = 0

z_{B^{'}} = z_{G} - B^{'} G \sin β = h - \frac{1}{3} m_{b} \sin β

z_{B^{'}} = \frac{1}{2} (z_{A} + z_{C})

z_{C} = 2 z_{B^{'}} - z_{A} = 2 h - \frac{2}{3} m_{b} \sin β - h - \frac{2}{3} m_{a} \sin α

\Rightarrow z_{C} = h - \frac{2}{3} (m_{a} \sin α + m_{b} \sin β)

z_{C} = h + \frac{2}{3} m_{c} \sin γ

\Rightarrow \frac{2}{3} m_{c} \sin γ = - \frac{2}{3} (m_{a} \sin α + m_{b} \sin β)

\Rightarrow \sin γ = - \frac{1}{g_{c}} (g_{a} \sin α + g_{b} \sin β)

\Rightarrow k {(g_{a} \sin α + g_{b} \sin β)}^{2} - (g_{a} \sin α + g_{b} \sin β) + h - {k g}_{c}^{2} = 0 \dots (i i i)

f r o m (i) a n d (i i) :

k (g_{a}^{2} \sin^{2} α - g_{b}^{2} \sin^{2} β) + (g_{a} \sin α - g_{b} \sin β) - k (g_{a}^{2} - g_{b}^{2}) = 0

(1 + {k g}_{a} \sin α + {k g}_{b} \sin β) (g_{a} \sin α - g_{b} \sin β) = k (g_{a}^{2} - g_{b}^{2})

f r o m (i) a n d (i i i) :

k [g_{a}^{2} \sin^{2} α - {(g_{a} \sin α + g_{b} \sin β)}^{2}] + [g_{a} \sin α + (g_{a} \sin α + g_{b} \sin β)] - k (g_{a}^{2} - g_{c}^{2}) = 0

(1 - {k g}_{b} \sin β) (2 g_{a} \sin α + g_{b} \sin β) = k (g_{a}^{2} - g_{c}^{2})

w e c a n s o l v e (n u m m e r i c a l l y)

{\begin{cases} (\frac{1}{k} + g_{a} \sin α + g_{b} \sin β) (g_{a} \sin α - g_{b} \sin β) = g_{a}^{2} - g_{b}^{2} \\ (\frac{1}{k} - g_{b} \sin β) (2 g_{a} \sin α + g_{b} \sin β) = g_{a}^{2} - g_{c}^{2} \end{cases}

f o r \sin α a n d \sin β .

n o t e : t h e r e c a n b e m o r e t h a n o n e

s o l u t i o n!

w i t h \sin α a n d \sin β w e g e t \sin γ f r o m

g_{a} \sin α + g_{b} \sin β + g_{c} \sin γ = 0

l e t e_{a}, e_{b}, e_{c} = p r o j e c t i o n o f A G, B G, C G

e_{a} = g_{a} \cos α = g_{a} \sqrt{1 - \sin^{2} α}

e_{b} = g_{b} \cos β = g_{b} \sqrt{1 - \sin^{2} β}

e_{c} = g_{c} \cos γ = g_{c} \sqrt{1 - \sin^{2} γ}

Δ_{P Q R} = \frac{3 \sqrt{(e_{a} + e_{b} + e_{c}) (- e_{a} + e_{b} + e_{c}) (e_{a} - e_{b} + e_{c}) (e_{a} + e_{b} - e_{c})}}{4}

Δ_{A B C} = \frac{\sqrt{(a + b + c) (- a + b + c) (a - b + c) (a + b - c)}}{4}

\frac{Δ_{P Q R}}{Δ_{A B C}} = \frac{3 \sqrt{(e_{a} + e_{b} + e_{c}) (- e_{a} + e_{b} + e_{c}) (e_{a} - e_{b} + e_{c}) (e_{a} + e_{b} - e_{c})}}{\sqrt{(a + b + c) (- a + b + c) (a - b + c) (a + b - c)}}

Commented by mr W last updated on 04/Dec/18

Commented by ajfour last updated on 06/Dec/18

T h a n k s s i r! a n d s o r r y f o r s u c h

n o t s o i n t e r e s t i n g q u e s t i o n . .

Commented by mr W last updated on 06/Dec/18

b u t i t i s a n i n t e r e s t i n g q u e s t i o n . w h a t

i s t h e r e a s o n f o r t h e r e s t r i c t i o n t h a t

t h e c e n t r o i d s h o u l d l i e o n z - a x i x ?

w i t h t h i s r e s t r i c t i o n n o t e v e r y t r i a n g l e

c a n b e p l a c e d i n s i d e t h e c u p, f o r

e x a m p l e a t r i a n g l e w i t h s i d e s 3, 4, 5 .

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